$\begin{aligned} &f(x)=x^2-6x+3 \\\\ &g(x)=\sqrt{x-2} \end{aligned}$ $f(g(11))=$
When evaluating composite functions, we work our way inside out. To evaluate $f(g(11))$, let's first evaluate $g(11)$. Then we'll plug that result into $f$ to find our answer. Let's evaluate $g({11})$. $\begin{aligned}g(x)&=\sqrt{x-2}\\\\ g({11})&=\sqrt{{11}-2} ~~~~~~~~~~\text{Plug in }x={11}\\\\ &=\sqrt{9}\\\\ &={3}\end{aligned}$ We now know that $f(g({11}))$ is the same as $f(3)$ because $g({11}) = 3$. Let's evaluate $f({3})$. $\begin{aligned}f(x)&=x^2 -6x + 3\\\\ f({{3}})&={3}^2 - 6({3}) + 3 ~~~~~~~~~~\text{Plug in }x={3}\\\\ &=9 - 18 + 3\\\\ &=-6\end{aligned}$ The answer: $f(g(11)) = -6$